Question

Consider the following distribution of daily wages of 50 workers of a factory

Daily wages (in Rs)	500 - 520	520 -540	540 - 560	560 - 580	580 -600
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer 1

To find the mean daily wages of the workers in the factory using the given frequency distribution, we can use the assumed mean method (also known as the step-deviation method). This method simplifies calculations, especially when dealing with grouped data. Here are the steps to calculate the mean daily wages:

1. Identify the midpoints of each class interval:
  - The midpoint (\(x_i\)) of a class interval is calculated as \(\frac{\text{Lower limit} + \text{Upper limit}}{2}\).

2. Calculate deviations from an assumed mean (\(A\)):
  - Choose an assumed mean (\(A\)) which is generally the midpoint of any class interval (preferably the one in the middle).
  - Calculate the deviation (\(d_i\)) of each midpoint from the assumed mean.

3. Calculate the frequency times deviation (\(f_i \cdot d_i\)) for each class interval.

4. Find the mean using the formula:
  \[\text{Mean} = A + \left( \frac{\sum f_i d_i}{\sum f_i} \right)\]
  where:
  - \(A\) is the assumed mean,
  - \(f_i\) is the frequency of the \(i\)-th class,
  - \(d_i\) is the deviation of the midpoint from the assumed mean,
  - \(\sum f_i\) is the sum of frequencies,
  - \(\sum f_i d_i\) is the sum of the product of frequencies and deviations.

Let's calculate the mean step by step:

Step 1: Calculate the midpoints (\(x_i\)) of each class interval
\[\begin{array}{cc}\text{Class Interval} & \text{Midpoint} (x_i) \\\hline500 - 520 & \frac{500 + 520}{2} = 510 \\520 - 540 & \frac{520 + 540}{2} = 530 \\540 - 560 & \frac{540 + 560}{2} = 550 \\560 - 580 & \frac{560 + 580}{2} = 570 \\580 - 600 & \frac{580 + 600}{2} = 590 \\\end{array}\]

Step 2: Choose an assumed mean (\(A\)) and calculate deviations (\(d_i\))
Let's choose \(A = 550\) (the midpoint of the third class).

\[\begin{array}{ccc}\text{Midpoint} (x_i) & \text{Deviation} (d_i = x_i - A) \\\hline510 & 510 - 550 = -40 \\530 & 530 - 550 = -20 \\550 & 550 - 550 = 0 \\570 & 570 - 550 = 20 \\590 & 590 - 550 = 40 \\\end{array}\]

 Step 3: Calculate the frequency times deviation (\(f_i \cdot d_i\))

\[\begin{array}{cccc}\text{Class Interval} & \text{Frequency} (f_i) & \text{Midpoint} (x_i) & f_i \cdot d_i \\\hline500 - 520 & 12 & 510 & 12 \times (-40) = -480 \\520 - 540 & 14 & 530 & 14 \times (-20) = -280 \\540 - 560 & 8 & 550 & 8 \times 0 = 0 \\560 - 580 & 6 & 570 & 6 \times 20 = 120 \\580 - 600 & 10 & 590 & 10 \times 40 = 400 \\\end{array}\]

Step 4: Calculate the sum of frequencies (\(\sum f_i\)) and the sum of frequency times deviation (\(\sum f_i d_i\))
\[\sum f_i = 12 + 14 + 8 + 6 + 10 = 50\]
\[\sum f_i d_i = -480 + (-280) + 0 + 120 + 400 = -240\]

 Step 5: Calculate the mean
\[\text{Mean} = A + \left( \frac{\sum f_i d_i}{\sum f_i} \right) = 550 + \left( \frac{-240}{50} \right) = 550 + (-4.8) = 545.2\]

Thus, the mean daily wages of the workers in the factory is Rs. 545.20.